$14.15$ Theorem: Suppose $F \in H(U-\{0\})$, $F$ is one-to-one in $U$, $F$ has a apole of order $1$ at $z=0$, with residue $1$ and neither $w_1$ nor $w_2$ are in $F(U$). Then $|w_1-w_2| \le 4$.
Proof: If $f=\frac1{F-w_1}$, then $f \in S$, hence $f(U) \supset D(0, \frac14)$, so the image of $U$ under $F-w_1$ contains all $w$ with $|w| > 4$. Since $w_2-w_1$ is not in this image, we have $|w_2 - w_1| \le 4$.
I understand only a part of the proof.
I know that the function $f = \frac1z + az + bz^2 + cz^3 + \cdots$ and then I show that $f$ is in class $S$ (defined at subsection). I used there Taylor's expantions of holomorphic function.
But I don't know how to prove that $w_1-w_2$ is not in image of $f-w_1$, and then how to prove the final result.
I'll be grateful for any explanations and help.
Suppose on the contrary that $w_2-w_1 \in (F-w_1)(U)=\{F(u)-w_1: u \in U\},$ then we have $w_2 \in F(U)$, which is a contradiction.
Hence $w_2-w_1 \notin (F-w_1)(U)$.
In the previous line, we have proved that if $|w| > 4 \implies w \in (F-w_1)(U)$.
Hence, the contrapositive is $w \notin (F-w_1)(U) \implies |w| \le 4.$
Since $w_2-w_1 \notin (F-w_1)(U)$, we conclude that $|w_2-w_1|\le 4$.